AREA is the quantity that expresses the extent of a two-dimensional
figure or shape , or planar lamina , in the plane .
The area of a shape can be measured by comparing the shape to squares
of a fixed size. In the
There are several well-known formulas for the areas of simple shapes such as triangles , rectangles , and circles . Using these formulas, the area of any polygon can be found by dividing the polygon into triangles . For shapes with curved boundary, calculus is usually required to compute the area. Indeed, the problem of determining the area of plane figures was a major motivation for the historical development of calculus . For a solid shape such as a sphere , cone , or cylinder , the area of its boundary surface is called the surface area . Formulas for the surface areas of simple shapes were computed by the ancient Greeks , but computing the surface area of a more complicated shape usually requires multivariable calculus .
CONTENTS * 1 Formal definition * 2 Units * 2.1 Conversions * 2.1.1 Non-metric units * 2.2 Other units including historical * 3 History * 3.1
* 4
* 4.1
* 4.1.1 Rectangles * 4.1.2 Dissection, parallelograms, and triangles * 4.2
* 4.2.1 Circles
* 4.2.2 Ellipses
* 4.2.3
* 4.3 General formulas * 4.3.1 Areas of 2-dimensional figures
* 4.3.2
* 4.4 List of formulas * 4.5 Relation of area to perimeter * 4.6 Fractals * 5
FORMAL DEFINITION See also:
An approach to defining what is meant by "area" is through axioms . "Area" can be defined as a function from a collection M of special kind of plane figures (termed measurable sets) to the set of real numbers which satisfies the following properties: * For all _S_ in _M_, _a_(_S_) ≥ 0. * If _S_ and _T_ are in _M_ then so are _S_ ∪ _T_ and _S_ ∩ _T_, and also _a_(_S_∪_T_) = _a_(_S_) + _a_(_T_) − _a_(_S_∩_T_). * If _S_ and _T_ are in _M_ with _S_ ⊆ _T_ then _T_ − _S_ is in _M_ and _a_(_T_−_S_) = _a_(_T_) − _a_(_S_). * If a set _S_ is in _M_ and _S_ is congruent to _T_ then _T_ is also in _M_ and _a_(_S_) = _a_(_T_). * Every rectangle _R_ is in _M_. If the rectangle has length _h_ and breadth _k_ then _a_(_R_) = _hk_. * Let _Q_ be a set enclosed between two step regions _S_ and _T_. A step region is formed from a finite union of adjacent rectangles resting on a common base, i.e. _S_ ⊆ _Q_ ⊆ _T_. If there is a unique number _c_ such that _a_(_S_) ≤ c ≤ _a_(_T_) for all such step regions _S_ and _T_, then _a_(_Q_) = _c_. It can be proved that such an area function actually exists. UNITS A square metre quadrat made of PVC pipe. Every unit of length has a corresponding unit of area, namely the area of a square with the given side length. Thus areas can be measured in square metres (m2), square centimetres (cm2), square millimetres (mm2), square kilometres (km2), square feet (ft2), square yards (yd2), square miles (mi2), and so forth. Algebraically, these units can be thought of as the squares of the corresponding length units. The SI unit of area is the square metre, which is considered an SI derived unit . CONVERSIONS Although there are 10 mm in 1 cm, there are 100 mm2 in 1 cm2. Calculation of the area of a square whose length and width are 1 metre would be: 1 metre x 1 metre = 1 m2 and therefore, another square with different sides can be calculated as: 3 metres x 2 metres = 6 m2. This is, however, equivalent to 6 million millimetres square. Following this, * 1 kilometre square = 1,000,000 metres square * 1 metre square= 10,000 centimetres square = 1,000,000 millimetres square * 1 centimetre square = 100 millimetres square Non-metric Units In non-metric units, the conversion between two square units is the square of the conversion between the corresponding length units. 1 foot = 12 inches , the relationship between square feet and square inches is 1 square foot = 144 square inches, where 144 = 122 = 12 × 12. Similarly: * 1 square yard = 9 square feet * 1 square mile = 3,097,600 square yards = 27,878,400 square feet In addition, conversion factors include: * 1 square inch = 6.4516 square centimetres * 1 square foot = 0.09290304 square metres * 1 square yard = 0.83612736 square metres * 1 square mile = 2.589988110336 square kilometres OTHER UNITS INCLUDING HISTORICAL See also: Category:Units of area There are several other common units for area. The "Are " was the original unit of area in the metric system , with; * 1 are = 100 square metres Though the are has fallen out of use, the hectare is still commonly used to measure land: * 1 hectare = 100 ares = 10,000 square metres = 0.01 square kilometres Other uncommon metric units of area include the tetrad , the hectad , and the myriad . The acre is also commonly used to measure land areas, where * 1 acre = 4,840 square yards = 43,560 square feet. An acre is approximately 40% of a hectare. On the atomic scale, area is measured in units of barns , such that: * 1 barn = 10−28 square meters. The barn is commonly used in describing the cross sectional area of interaction in nuclear physics . In India, * 20 Dhurki = 1 Dhur * 20 Dhur = 1 Khatha * 20 Khata = 1 Bigha * 32 Khata = 1 Acre HISTORY CIRCLE AREA In the 5th century BCE,
Subsequently, Book I of Euclid\'s _Elements_ dealt with equality of
areas between two-dimensional figures. The mathematician Archimedes
used the tools of
Swiss scientist
TRIANGLE AREA Heron (or Hero) of Alexandria found what is known as Heron\'s formula
for the area of a triangle in terms of its sides, and a proof can be
found in his book, _Metrica_, written around 60 CE. It has been
suggested that
In 499
A formula equivalent to Heron's was discovered by the Chinese
independently of the Greeks. It was published in 1247 in _Shushu
Jiuzhang_ ("
QUADRILATERAL AREA In the 7th century CE,
GENERAL POLYGON AREA The development of Cartesian coordinates by
AREAS DETERMINED USING CALCULUS The development of integral calculus in the late 17th century provided tools that could subsequently be used for computing more complicated areas, such as the area of an ellipse and the surface areas of various curved three-dimensional objects. AREA FORMULAS POLYGON FORMULAS Main article:
For a non-self-intersecting (simple ) polygon, the Cartesian coordinates ( x i , y i ) {displaystyle (x_{i},y_{i})} _ (i_=0, 1, ..., _n_-1) of whose _n_ vertices are known, the area is given by the surveyor\'s formula : A = 1 2 i = 0 n 1 ( x i y i + 1 x i + 1 y i ) , {displaystyle A={frac {1}{2}}sum _{i=0}^{n-1}(x_{i}y_{i+1}-x_{i+1}y_{i}),} where when _i_=_n_-1, then _i_+1 is expressed as modulus _n_ and so refers to 0. Rectangles The area of this rectangle is lw. The most basic area formula is the formula for the area of a rectangle . Given a rectangle with length l and width w, the formula for the area is: _A_ = _lw_ (rectangle). That is, the area of the rectangle is the length multiplied by the width. As a special case, as _l_ = _w_ in the case of a square, the area of a square with side length s is given by the formula: _A_ = _s_2 (square). The formula for the area of a rectangle follows directly from the basic properties of area, and is sometimes taken as a definition or axiom . On the other hand, if geometry is developed before arithmetic , this formula can be used to define multiplication of real numbers . Equal area figures. Dissection, Parallelograms, And Triangles Main articles:
Most other simple formulas for area follow from the method of dissection . This involves cutting a shape into pieces, whose areas must sum to the area of the original shape. For an example, any parallelogram can be subdivided into a trapezoid and a right triangle , as shown in figure to the left. If the triangle is moved to the other side of the trapezoid, then the resulting figure is a rectangle. It follows that the area of the parallelogram is the same as the area of the rectangle: _A_ = _bh_ (parallelogram). Two equal triangles. However, the same parallelogram can also be cut along a diagonal into two congruent triangles, as shown in the figure to the right. It follows that the area of each triangle is half the area of the parallelogram: A = 1 2 b h {displaystyle A={frac {1}{2}}bh} (triangle). Similar arguments can be used to find area formulas for the trapezoid as well as more complicated polygons . AREA OF CURVED SHAPES Circles A circle can be divided into sectors which rearrange to form an
approximate parallelogram . Main article:
The formula for the area of a circle (more properly called the area enclosed by a circle or the area of a disk ) is based on a similar method. Given a circle of radius _r_, it is possible to partition the circle into sectors , as shown in the figure to the right. Each sector is approximately triangular in shape, and the sectors can be rearranged to form and approximate parallelogram. The height of this parallelogram is _r_, and the width is half the circumference of the circle, or π_r_. Thus, the total area of the circle is π_r_2: _A_ = π_r_2 (circle). Though the dissection used in this formula is only approximate, the error becomes smaller and smaller as the circle is partitioned into more and more sectors. The limit of the areas of the approximate parallelograms is exactly π_r_2, which is the area of the circle. This argument is actually a simple application of the ideas of calculus . In ancient times, the method of exhaustion was used in a similar way to find the area of the circle, and this method is now recognized as a precursor to integral calculus . Using modern methods, the area of a circle can be computed using a definite integral : A = 2 r r r 2 x 2 d x = r 2 . {displaystyle A;=;2int _{-r}^{r}{sqrt {r^{2}-x^{2}}},dx;=;pi r^{2}.} Ellipses Main article:
The formula for the area enclosed by an ellipse is related to the formula of a circle; for an ellipse with semi-major and semi-minor axes _x_ and _y_ the formula is: A = x y . {displaystyle A=pi xy.} Surface Area Main article:
Most basic formulas for surface area can be obtained by cutting surfaces and flattening them out. For example, if the side surface of a cylinder (or any prism ) is cut lengthwise, the surface can be flattened out into a rectangle. Similarly, if a cut is made along the side of a cone , the side surface can be flattened out into a sector of a circle, and the resulting area computed. The formula for the surface area of a sphere is more difficult to
derive: because a sphere has nonzero
where _r_ is the radius of the sphere. As with the formula for the area of a circle, any derivation of this formula inherently uses methods similar to calculus . GENERAL FORMULAS Areas Of 2-dimensional Figures * A triangle : 1 2 B h {displaystyle {tfrac {1}{2}}Bh} _ (where B_ is any side, and _h_ is the distance from the line on which _B_ lies to the other vertex of the triangle). This formula can be used if the height _h_ is known. If the lengths of the three sides are known then _Heron\'s formula _ can be used: s ( s a ) ( s b ) ( s c ) {displaystyle {sqrt {s(s-a)(s-b)(s-c)}}} _ where a_, _b_, _c_ are the sides of the triangle, and s = 1 2 ( a + b + c ) {displaystyle s={tfrac {1}{2}}(a+b+c)} _ is half of its perimeter. If an angle and its two included sides are given, the area is 1 2 a b sin ( C ) {displaystyle {tfrac {1}{2}}absin(C)} where C_ is the given angle and _a_ and _b_ are its included sides. If the triangle is graphed on a coordinate plane, a matrix can be used and is simplified to the absolute value of 1 2 ( x 1 y 2 + x 2 y 3 + x 3 y 1 x 2 y 1 x 3 y 2 x 1 y 3 ) {displaystyle {tfrac {1}{2}}(x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1}-x_{2}y_{1}-x_{3}y_{2}-x_{1}y_{3})} _. This formula is also known as the shoelace formula and is an easy way to solve for the area of a coordinate triangle by substituting the 3 points (x1,y1)_, _(x2,y2)_, and _(x3,y3)_. The shoelace formula can also be used to find the areas of other polygons when their vertices are known. Another approach for a coordinate triangle is to use calculus to find the area. * A simple polygon constructed on a grid of equal-distanced points (i.e., points with integer coordinates) such that all the polygon's vertices are grid points: i + b 2 1 {displaystyle i+{frac {b}{2}}-1} _, where i_ is the number of grid points inside the polygon and _b_ is the number of boundary points. This result is known as Pick\'s theorem .
_ Integration can be thought of as measuring the area under a curve, defined by f_(_x_), between two points (here _a_ and _b_). _ The area between two graphs can be evaluated by calculating the difference between the integrals of the two functions * The area between a positive-valued curve and the horizontal axis, measured between two values a_ and _b_ (b is defined as the larger of the two values) on the horizontal axis, is given by the integral from _a_ to _b_ of the function that represents the curve: A = a b f ( x ) d x . {displaystyle A=int _{a}^{b}f(x),dx.} _ * The area between the graphs of two functions is equal to the integral of one function , f_(_x_), minus the integral of the other function, _g_(_x_): A = a b ( f ( x ) g ( x ) ) d x , {displaystyle A=int _{a}^{b}(f(x)-g(x)),dx,} _ where f ( x ) {displaystyle f(x)} is the curve with the greater y-value. * An area bounded by a function r_ = _r_(θ) expressed in polar coordinates is: A = 1 2 r 2 d . {displaystyle A={1 over 2}int r^{2},dtheta .} * The area enclosed by a parametric curve u ( t ) = ( x ( t ) , y ( t ) ) {displaystyle {vec {u}}(t)=(x(t),y(t))} with endpoints u ( t 0 ) = u ( t 1 ) {displaystyle {vec {u}}(t_{0})={vec {u}}(t_{1})} is given by the line integrals : t 0 t 1 x y d t = t 0 t 1 y x d t = 1 2 t 0 t 1 ( x y y x ) d t {displaystyle oint _{t_{0}}^{t_{1}}x{dot {y}},dt=-oint _{t_{0}}^{t_{1}}y{dot {x}},dt={1 over 2}oint _{t_{0}}^{t_{1}}(x{dot {y}}-y{dot {x}}),dt} (see Green\'s theorem ) or the _z_-component of 1 2 t 0 t 1 u u d t . {displaystyle {1 over 2}oint _{t_{0}}^{t_{1}}{vec {u}}times {dot {vec {u}}},dt.} Bounded
To find the bounded area between two quadratic functions , we subtract one from the other to write the difference as f ( x ) g ( x ) = a x 2 + b x + c = a ( x ) ( x ) {displaystyle f(x)-g(x)=ax^{2}+bx+c=a(x-alpha )(x-beta )} where _f_(_x_) is the quadratic upper bound and _g_(_x_) is the quadratic lower bound. Define the discriminant of _f_(_x_)-_g_(_x_) as = b 2 4 a c . {displaystyle Delta =b^{2}-4ac.} By simplifying the integral formula between the graphs of two functions (as given in the section above) and using Vieta\'s formula , we can obtain A = 6 a 2 = a 6 ( ) 3 , a 0. {displaystyle A={frac {Delta {sqrt {Delta }}}{6a^{2}}}={frac {a}{6}}(beta -alpha )^{3},qquad aneq 0.} The above remains valid if one of the bounding functions is linear instead of quadratic. Surface
*
General
The general formula for the surface area of the graph of a continuously differentiable function z = f ( x , y ) , {displaystyle z=f(x,y),} where ( x , y ) D R 2 {displaystyle (x,y)in Dsubset mathbb {R} ^{2}} and D {displaystyle D} is a region in the xy-plane with the smooth boundary: A = D ( f x ) 2 + ( f y ) 2 + 1 d x d y . {displaystyle A=iint _{D}{sqrt {left({frac {partial f}{partial x}}right)^{2}+left({frac {partial f}{partial y}}right)^{2}+1}},dx,dy.} An even more general formula for the area of the graph of a parametric surface in the vector form r = r ( u , v ) , {displaystyle mathbf {r} =mathbf {r} (u,v),} where r {displaystyle mathbf {r} } is a continuously differentiable vector function of ( u , v ) D R 2 {displaystyle (u,v)in Dsubset mathbb {R} ^{2}} is: A = D r u r v d u d v . {displaystyle A=iint _{D}left{frac {partial mathbf {r} }{partial u}}times {frac {partial mathbf {r} }{partial v}}right,du,dv.} LIST OF FORMULAS Additional common formulas for area: SHAPE FORMULA VARIABLES Regular triangle (equilateral triangle ) 3 4 s 2 {displaystyle {frac {sqrt {3}}{4}}s^{2},!} s {displaystyle s} is the length of one side of the triangle.
Regular hexagon 3 2 3 s 2 {displaystyle {frac {3}{2}}{sqrt {3}}s^{2},!} s {displaystyle s} is the length of one side of the hexagon. Regular octagon 2 ( 1 + 2 ) s 2 {displaystyle 2(1+{sqrt {2}})s^{2},!} s {displaystyle s} is the length of one side of the octagon.
Regular polygon 1 4 n p 2 cot ( / n ) {displaystyle {frac {1}{4n}}p^{2}cdot cot(pi /n),!} p {displaystyle p} is the perimeter and n {displaystyle n} is the number of sides. Regular polygon 1 2 n R 2 sin ( 2 / n ) = n r 2 tan ( / n ) {displaystyle {frac {1}{2}}nR^{2}cdot sin(2pi /n)=nr^{2}tan(pi /n),!} R {displaystyle R} is the radius of a circumscribed circle, r {displaystyle r} is the radius of an inscribed circle, and n {displaystyle n} is the number of sides. Regular polygon 1 2 a p = 1 2 n s a {displaystyle {tfrac {1}{2}}ap={tfrac {1}{2}}nsa,!} n {displaystyle n} is the number of sides, s {displaystyle s} is the side length, a {displaystyle a} is the apothem , or the radius of an inscribed circle in the polygon, and p {displaystyle p} is the perimeter of the polygon.
Total surface area of a cylinder 2 r ( r + h ) {displaystyle 2pi r(r+h),!} r {displaystyle r} and h {displaystyle h} are the radius and height, respectively. Lateral surface area of a cylinder 2 r h {displaystyle 2pi rh,!} r {displaystyle r} and h {displaystyle h} are the radius and height, respectively. Total surface area of a sphere 4 r 2 or d 2 {displaystyle 4pi r^{2} {text{or}} pi d^{2},!} r {displaystyle r} and d {displaystyle d} are the radius and diameter, respectively. Total surface area of a pyramid B + P L 2 {displaystyle B+{frac {PL}{2}},!} B {displaystyle B} is the base area, P {displaystyle P} is the base perimeter and L {displaystyle L} is the slant height. Total surface area of a pyramid frustum B + P L 2 {displaystyle B+{frac {PL}{2}},!} B {displaystyle B} is the base area, P {displaystyle P} is the base perimeter and L {displaystyle L} is the slant height.
Circular to square area conversion 4 C {displaystyle {frac {pi }{4}}C,!} C {displaystyle C} is the area of the circle in circular units. The above calculations show how to find the areas of many common shapes . The areas of irregular polygons can be calculated using the "Surveyor\'s formula ". RELATION OF AREA TO PERIMETER The isoperimetric inequality states that, for a closed curve of length _L_ (so the region it encloses has perimeter _L_) and for area _A_ of the region that it encloses, 4 A L 2 , {displaystyle 4pi Aleq L^{2},} and equality holds if and only if the curve is a circle . Thus a circle has the largest area of any closed figure with a given perimeter. At the other extreme, a figure with given perimeter _L_ could have an arbitrarily small area, as illustrated by a rhombus that is "tipped over" arbitrarily far so that two of its angles are arbitrarily close to 0° and the other two are arbitrarily close to 180°. For a circle, the ratio of the area to the circumference (the term for the perimeter of a circle) equals half the radius _r_. This can be seen from the area formula _πr_2 and the circumference formula 2_πr_. The area of a regular polygon is half its perimeter times the apothem (where the apothem is the distance from the center to the nearest point on any side). FRACTALS Doubling the edge lengths of a polygon multiplies its area by four, which is two (the ratio of the new to the old side length) raised to the power of two (the dimension of the space the polygon resides in). But if the one-dimensional lengths of a fractal drawn in two dimensions are all doubled, the spatial content of the fractal scales by a power of two that is not necessarily an integer. This power is called the fractal dimension of the fractal. AREA BISECTORS Main article: Bisection §
There are an infinitude of lines that bisect the area of a triangle. Three of them are the medians of the triangle (which connect the sides' midpoints with the opposite vertices), and these are concurrent at the triangle's centroid ; indeed, they are the only area bisectors that go through the centroid. Any line through a triangle that splits both the triangle's area and its perimeter in half goes through the triangle's incenter (the center of its incircle ). There are either one, two, or three of these for any given triangle. Any line through the midpoint of a parallelogram bisects the area. All area bisectors of a circle or other ellipse go through the center, and any chords through the center bisect the area. In the case of a circle they are the diameters of the circle. OPTIMIZATION Given a wire contour, the surface of least area spanning ("filling") it is a minimal surface . Familiar examples include soap bubbles . The question of the filling area of the
The circle has the largest area of any two-dimensional object having the same perimeter. A cyclic polygon (one inscribed in a circle) has the largest area of any polygon with a given number of sides of the same lengths. A version of the isoperimetric inequality for triangles states that the triangle of greatest area among all those with a given perimeter is equilateral . The triangle of largest area of all those inscribed in a given circle is equilateral; and the triangle of smallest area of all those circumscribed around a given circle is equilateral. The ratio of the area of the incircle to the area of an equilateral triangle, 3 3 {displaystyle {frac {pi }{3{sqrt {3}}}}} , is larger than that of any non-equilateral triangle. The ratio of the area to the square of the perimeter of an equilateral triangle, 1 12 3 , {displaystyle {frac {1}{12{sqrt {3}}}},} is larger than that for any other triangle. SEE ALSO *
* Routh\'s theorem , a generalization of the one-seventh area triangle. *
REFERENCES * ^ _A_ _B_ _C_ _D_ _E_ _F_ _G_ _H_ Weisstein, Eric W. "Area".
EXTERNAL LINKS Find more aboutAREAat's siste |